A more Intuitive proof of regularity of topological group [closed]












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Does someone knows a good reference for the following result?



"A topological group is Hausdorff if and only if the identity is closed."



I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.










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closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
    $endgroup$
    – Randall
    Dec 12 '18 at 12:17










  • $begingroup$
    Maybe you can list some because I can't even find a single Wikipedia page for it
    $endgroup$
    – YuiTo Cheng
    Dec 12 '18 at 12:21
















0












$begingroup$


Does someone knows a good reference for the following result?



"A topological group is Hausdorff if and only if the identity is closed."



I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.










share|cite|improve this question











$endgroup$



closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
    $endgroup$
    – Randall
    Dec 12 '18 at 12:17










  • $begingroup$
    Maybe you can list some because I can't even find a single Wikipedia page for it
    $endgroup$
    – YuiTo Cheng
    Dec 12 '18 at 12:21














0












0








0





$begingroup$


Does someone knows a good reference for the following result?



"A topological group is Hausdorff if and only if the identity is closed."



I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.










share|cite|improve this question











$endgroup$




Does someone knows a good reference for the following result?



"A topological group is Hausdorff if and only if the identity is closed."



I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.







general-topology topological-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 9:21







YuiTo Cheng

















asked Dec 12 '18 at 11:46









YuiTo ChengYuiTo Cheng

1,325526




1,325526




closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
    $endgroup$
    – Randall
    Dec 12 '18 at 12:17










  • $begingroup$
    Maybe you can list some because I can't even find a single Wikipedia page for it
    $endgroup$
    – YuiTo Cheng
    Dec 12 '18 at 12:21














  • 2




    $begingroup$
    I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
    $endgroup$
    – Randall
    Dec 12 '18 at 12:17










  • $begingroup$
    Maybe you can list some because I can't even find a single Wikipedia page for it
    $endgroup$
    – YuiTo Cheng
    Dec 12 '18 at 12:21








2




2




$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17




$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17












$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21




$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21










1 Answer
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oldest

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$begingroup$

Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.



Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.



This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.



The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.



    Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.



    This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.



    The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.



      Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.



      This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.



      The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.



        Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.



        This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.



        The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.






        share|cite|improve this answer









        $endgroup$



        Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.



        Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.



        This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.



        The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 12:04









        CronusCronus

        1,098518




        1,098518















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